The inductor doesn't matter for average current.
Average means DC, and with DC all you need is the average (DC) voltage.

For a half-wave rectified voltage the DC value is , or 31.8 V.
If you subtract the diode drop and divide by the 100 ohm resistance, you get 310 mA.

Impedance calculations don't make sense because they only work for steady state sinusoidal circuits. The diode is non-linear. What you could do is calculate the fourier transform of a half wave rectified voltage and then do impedance calculations but you'd need to do it for a lot of harmonics...

The inductor doesn't matter for average current.
Average means DC, and with DC all you need is the average (DC) voltage.

For a half-wave rectified voltage the DC value is , or 31.8 V.
If you subtract the diode drop and divide by the 100 ohm resistance, you get 310 mA.

Impedance calculations don't make sense because they only work for steady state sinusoidal circuits. The diode is non-linear. What you could do is calculate the fourier transform of a half wave rectified voltage and then do impedance calculations but you'd need to do it for a lot of harmonics...

Click to expand...

It seems the issue in this case is to do with the circuit transient behavior during the circuit (diode) conduction / non-conduction intervals. The time constant is 150usec which is short compared with the base AC period of 2.5msec. Each current pulse is therefore a composite of a sinusoid and an exponentially decaying term. It seems credible that one might solve for the transient expression of the current pulse and then determine the mean of the pulse, taking into consideration the pulse occurs over a yet to be determined interval. If the inductor were sufficiently large to give continuous current then the solution will probably (?) be different.

Like you, I initially thought it was just a trivial exercise and then I took a closer look. I'm thinking of doing a simulation to see what the real story is. I'm not sure whether all this work is really worth it. A "ball park" approach will probably do.

The inductor doesn't matter for average current.
Average means DC, and with DC all you need is the average (DC) voltage.

For a half-wave rectified voltage the DC value is , or 31.8 V.
If you subtract the diode drop and divide by the 100 ohm resistance, you get 310 mA.

Impedance calculations don't make sense because they only work for steady state sinusoidal circuits. The diode is non-linear. What you could do is calculate the fourier transform of a half wave rectified voltage and then do impedance calculations but you'd need to do it for a lot of harmonics...